How to force dispatch to an internal generic in R? - r

I have a class 'myClass' in R that is essentially a list. It has an assignment operator which is going to do some things and then should assign the value using the regular list assignment operator
`$<-.myClass`<-function(x,i,value){
# do some pre-processing stuff
# make the assignment using the default list assignment
x[[i]]<-value
x
}
But I can't actually use x[[i]]<-value as it will dispatch to the already existing [[<-.myClass method.
In similar S3 dispatching cases, I've been able use UseMethod or specifically call [[<-.list, or [[<-.default but those don't seem to exist because $<- and [[<- are primitive generics, right? And I'm sure I'll be sent to a special R hell if I try to call .Primitive("$<-"). What is the correct way to dispatch the assignment to the default assignment method?

It doesn't look like there is a particularly elegant way to do this. The data.frame method for $<- looks like this:
`$<-.data.frame` <- function (x, name, value) {
cl <- oldClass(x)
class(x) <- NULL
x[[name]] <- value
class(x) <- cl
x
}
(with error checking code omitted)
This should only create one copy of x, because class<- modifies in place, and so does the default method for [[<-.

Related

rlang: Error: Can't convert a function to a string

I created a function to convert a function name to string. Version 1 func_to_string1 works well, but version 2 func_to_string2 doesn't work.
func_to_string1 <- function(fun){
print(rlang::as_string(rlang::enexpr(fun)))
}
func_to_string2 <- function(fun){
is.function(fun)
print(rlang::as_string(rlang::enexpr(fun)))
}
func_to_string1 works:
> func_to_string1(sum)
[1] "sum"
func_to_string2 doesn't work.
> func_to_string2(sum)
Error: Can't convert a primitive function to a string
Call `rlang::last_error()` to see a backtrace
My guess is that by calling the fun before converting it to a string, it gets evaluated inside function and hence throw the error message. But why does this happen since I didn't do any assignments?
My questions are why does it happen and is there a better way to convert function name to string?
Any help is appreciated, thanks!
This isn't a complete answer, but I don't think it fits in a comment.
R has a mechanism called pass-by-promise,
whereby a function's formal arguments are lazy objects (promises) that only get evaluated when they are used.
Even if you didn't perform any assignment,
the call to is.function uses the argument,
so the promise is "replaced" by the result of evaluating it.
Nevertheless, in my opinion, this seems like an inconsistency in rlang*,
especially given cory's answer,
which implies that R can still find the promise object even after a given parameter has been used;
the mechanism to do so might not be part of R's public API though.
*EDIT: see coments.
Regardless, you could treat enexpr/enquo/ensym like base::missing,
in the sense that you should only use them with parameters you haven't used at all in the function's body.
Maybe use this instead?
func_to_string2 <- function(fun){
is.function(fun)
deparse(substitute(fun))
#print(rlang::as_string(rlang::enexpr(fun)))
}
> func_to_string2(sum)
[1] "sum"
This question brings up an interesting point on lazy evaluations.
R arguments are lazily evaluated, meaning the arguments are not evaluated until its required.
This is best understood in the Advanced R book which has the following example,
f <- function(x) {
10
}
f(stop("This is an error!"))
the result is 10, which is surprising because x is never called and hence never evaluated. We can force x to be evaluated by using force()
f <- function(x) {
force(x)
10
}
f(stop("This is an error!"))
This behaves as expected. In fact we dont even need force() (Although it is good to be explicit).
f <- function(x) {
x
10
}
f(stop("This is an error!"))
This what is happening with your call here. The function sum which is a symbol initially is being evaluated with no arguments when is.function() is being called. In fact, even this will fail.
func_to_string2 <- function(fun){
fun
print(rlang::as_string(rlang::ensym(fun)))
}
Overall, I think its best to use enexpr() at the very beginning of the function.
Source:
http://adv-r.had.co.nz/Functions.html

Call Arguments of Function inside Function / R language

I have a function:
func <- function (x)
{
arguments <- match.call()
return(arguments)
}
1) If I call my function with specifying argument in the call:
func("value")
I get:
func(x = "value")
2) If I call my function by passing a variable:
my_variable <-"value"
func(my_variable)
I get:
func(x = my_variable)
Why is the first and the second result different?
Can I somehow get in the second call "func(x = "value")"?
I'm thinking my problem is that the Environment inside a function simply doesn't contain values if they were passed by variables. The Environment contains only names of variables for further lookup. Is there a way to follow such reference and get value from inside a function?
In R, when you pass my_variable as formal argument x into a function, the value of my_variable will only be retrieved when the function tries to read x (if it does not use x, my_variable will not be read at all). The same applies when you pass more complicated arguments, such as func(x = compute_my_variable()) -- the call to compute_my_variable will take place when func tries to read x (this is referred to as lazy evaluation).
Given lazy evaluation, what you are trying to do is not well defined because of side effects - in which order would you like to evaluate the arguments? Which arguments would you like to evaluate at all? (note a function can just take an expression for its argument using substitute, but not evaluate it). As a side effect, compute_my_variable could modify something that would impact the result of another argument of func. This can happen even when you only passed variables and constants as arguments (function func could modify some of the variables that will be later read, or even reading a variable such as my_variable could trigger code that would modify some of the variables that will be read later, e.g. with active bindings or delayed assignment).
So, if all you want to do is to log how a function was called, you can use sys.call (or match.call but that indeed expands argument names, etc). If you wanted a more complete stacktrace, you can use e.g. traceback(1).
If for some reason you really wanted values of all arguments, say as if they were all read in the order of match.call, which is the order in which they are declared, you can do it using eval (returns them as list):
lapply(as.list(match.call())[-1], eval)
can't you simply
return paste('func(x =', x, ')')

Overloading R function - is this right?

consumeSingleRequest <- function(api_key, URL, columnNames, globalParam="", ...)
consumeSingleRequest <- function(api_key, URL, columnNames, valuesList, globalParam="")
I am trying to overload a function like this, that takes in multiple lists in the first function and combines them into one list of lists. However, I don't seem to be able to skip passing in globalParam and pass in oly the multiple lists in the ...
Does anyone know how to do that?
I've heard S3 methods could be used for that? Does anyone know how?
R doesn't support a concept of overloading functions. It supports function calls with variable number of arguments. So you can declare a function with any number of arguments, but supply only a subset of those when actually calling a function. Take vector function as an example:
> vector
function (mode = "logical", length = 0L)
.Internal(vector(mode, length))
<bytecode: 0x103b89070>
<environment: namespace:base>
It supports up to 2 parameters, but can be called with none or some subset(in that case default values are used) :
> vector()
logical(0)
> vector(mode='numeric')
numeric(0)
So you only need a second declaration:
consumeSingleRequest <- function(api_key, URL, columnNames, valuesList, globalParam="")
And supply just supply the needed parameters when actually calling the function
consumeSingleRequest(api_key=..., valueList=...)
P.S. A good explanation can be found in Advanced R Book.

How to use reflection to intercept an expression prior to evaluation?

I was hoping to use R's reflection capabilities to intercept the current expression under evaluation before it is evaluated.
For instance, to create some syntax sugar, given the following:
> Server <- setRefClass("Server",
> methods = list(
> handler = function(expr) submitExpressionToRemoteServer(expr)
> )
> )
> server <- Server()
> server$foo$bar$baz #... should be map to... server$handler("foo$bar$baz")
I want the expression server$foo$bar$baz to be intercepted by the server$handlermethod and get mapped to server$handler("foo$bar$baz").
Note that I want this call to succeed even though server$foo is not defined: I am interested only in the expression itself (so I can do stuff with the expression), not that it evaluates to a valid local object.
Is this possible?
I don't think this is possible to redefine the $ behavior with Reference Classes (R5) objects in R. However, this is something that you can do with S4 classes. The main problem is that an expression like
server$foo$bar$baz
would get translated to a series of calls like
$($($(server,"foo"),"bar"),"baz")
but unlike normal function nesting, each inner call appears to be fully evaluated before going to the next level of nesting. This it's not really possible just to split up everything after the first $ because that's not how it's parsed. However you can have the $ function return another object and append all the values sent to the object. Here's a sample S4 class
setClass("Server", slots=list(el="character"))
setMethod("$", signature(x="Server"),
function(x,name) {
xx <- append(slot(x,"el"),name)
new("Server", el=xx)
}
)
server <- new("Server")
server$foo$bar$baz
# An object of class "Server"
# Slot "el":
# [1] "foo" "bar" "baz"
the only problem is there's no way i've found to know when you're at the end of a list if you wanted to do anything with those parameters.

Can an R function access its own name?

Can you write a function that prints out its own name?
(without hard-coding it in, obviously)
You sure can.
fun <- function(x, y, z) deparse(match.call()[[1]])
fun(1,2,3)
# [1] "fun"
You can, but just in case it's because you want to call the function recursively see ?Recall which is robust to name changes and avoids the need to otherwise process to get the name.
Recall package:base R Documentation
Recursive Calling
Description:
‘Recall’ is used as a placeholder for the name of the function in
which it is called. It allows the definition of recursive
functions which still work after being renamed, see example below.
As you've seen in the other great answers here, the answer seems to be "yes"...
However, the correct answer is actually "yes, but not always". What you can get is actually the name (or expression!) that was used to call the function.
First, using sys.call is probably the most direct way of finding the name, but then you need to coerce it into a string. deparse is more robust for that.
myfunc <- function(x, y=42) deparse(sys.call()[[1]])
myfunc (3) # "myfunc"
...but you can call a function in many ways:
lapply(1:2, myfunc) # "FUN"
Map(myfunc, 1:2) # (the whole function definition!)
x<-myfunc; x(3) # "x"
get("myfunc")(3) # "get(\"myfunc\")"
The basic issue is that a function doesn't have a name - it's just that you typically assign the function to a variable name. Not that you have to - you can have anonymous functions - or assign many variable names to the same function (the x case above).

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